#$&* course mth 158 023. `* 23
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Given Solution: * * The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6). The function then decreases to its local min at (5, 0), after which it continues increasing. So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.3.22 (was 3.2.12). Piecewise linear (-3,3) to (-1,0) to (0,2) to (1,0) to (3,3).Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither. Your solution: Intersects the x axis at (-1, 0) and (1, 0) and the y axis at (0, 2) decreases from (-3,3) to (-1,0) at the interval (-3, -1) decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1) increases from (-1,0) to (0, 2) on the interval (-1, 0) increases from (1,0) to (3, 3) on the interval (1, 3) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The function intersects the x axis at (-1, 0) and (1, 0), and the y axis at (0, 2). The function decreases from (-3,3) to (-1,0) so it is decreasing on the interval (-3, -1). The function decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1). The function increases from (-1,0) to (0, 2) so it is increasing on the interval (-1, 0). The function ioncreases from (1,0) to (3, 3) so it is increasing on the interval (1, 3). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is 0 <= y <= 3, written as the interval [0, 3]. The function is symmetric about the y axis, with f(-x) = f(x) for every x (e.g., f(-3) = f(3) = 3; f(-1) = f(1) = 0; etc.). So the function is even. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.3.28 (was 3.2.18). Piecewise linear (-3,-2) to (-2, 1) to (0, 1) to (2, 2) to (3, 0)Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Intersects x axis at (-2.25, 0) and (3, 0) and y axis at (0, 1) decreases from (2,2) to (3,0) increases from (-3,-2) to (-2, 1) increases from (0, 1) to (2, 2) function is constant on (-2, 0) the function is not even confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The function intersects the x axis at (-2.25, 0) and (3, 0), and the y axis at (0, 1). The function decreases from (2,2) to (3,0) so it is decreasing on the interval (2,3). The function increases from (-3,-2) to (-2, 1) so it is increasing on the interval (-3, -2). The function ioncreases from (0, 1) to (2, 2) so it is increasing on the interval (0, 2). The function value does not change between (-2, 1) and (0, 1) so the function is constant on (-2, 0). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is -2 <= y <= 2, written as the interval [-2, 2]. The function is not symmetric about the y axis, with f(-x) not equal to f(x) for many values of x; e.g., f(-3) is -2 and f(3) is 0. So the function is not even. x = 3 shows us that the function is not odd either, since for an odd function f(-x) = -f(x) and for x = 3 this is not the case. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.3.32 (was 3.2.24). sine-type fn (-pi,-1) to (0, 2) to (pi, -1).At what numbers does the function have a local max and what are these local maxima?At what numbers does the function have a local min and what are these local minima? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Maximum (0,1) and minimum are (-pi, 1) and (pi, -1) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Local maximum is (0,1) Local minimum are (-pi,-1) and (pi,-1) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.3.76 / 46 (was 3.2.36) (f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2 What is your expression for (f(x) - f(1) ) / (x - 1) and how did you get this expression? How did you use your result to get the ave rate of change from x = 1 to x = 2, and what is your value? What is the equation of the secant line from the x = 1 point to the x = 2 point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) - f(1) = (x-2x^2) - (-1) = -2 x^2 + x + 1. (2x + 1) ( -x + 1) (f(x) - f(1) ) / ( x - 1) = (2x + 1) ( -x + 1) / (x - 1) = (2x + 1) * -1 = - (2x + 1) x = 2 ( f(2) - f(1) ) / ( 2 - 1) slope: (1, f(1) ) to (2, f(2)) -(2 * 2 + 1) = -5 secant line: (y - (-1) ) = -5 * (x - 1) y = -5 x + 4 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * f(x) - f(1) = (x-2x^2) - (-1) = -2 x^2 + x + 1. This factors into (2x + 1) ( -x + 1). Since -x + 1 = - ( x - 1) we obtain (f(x) - f(1) ) / ( x - 1) = (2x + 1) ( -x + 1) / (x - 1) = (2x + 1) * -1 = - (2x + 1). A secant line runs from one point of the graph to another. The secant line here runs from the graph point (1, f(1) ) to the graph ponit ( x , f(x) ), and the expression we have just obtained is the slope of the secant line. For x = 2 the expression -(2x + 1) gives us ( f(2) - f(1) ) / ( 2 - 1), which is the slope of the line from (1, f(1) ) to (2, f(2)) . -(2 * 2 + 1) = -5, which is the desired slope. The secant line contains the point (1, -1) and has slope 5. So the equation of the secant line is (y - (-1) ) = -5 * (x - 1), which we solve to obtain y = -5 x + 4. ** STUDENT QUESTION 100724 I'm OK up to the point where ( f(x) - f(1) ) / (x - 1) = - ( 2 x + 1 ). Beyond that, I don't know where to go and can't follow the given solution. INSTRUCTOR RESPONSE You appear to understand the expression (f(x) - f(1) ) / (x - 1), and you appear to know how to simplify the resulting numerator, factor, and simplify the resulting expression. That's a very good start. Optional series of exercises: Here's a numerical exercise, using the same function. What is numerical value of the expression (f(3) - f(1) ) / (3 - 1)? &&&&What are the x = 1 and x = 3 points of a graph of f(x) vs. x?&&&& Sketch a set of coordinate axes, and plot these two points. Connect the two points with a straight line segment. Describe your sketch. &&&& What is the rise from the first point to the second? &&&&What is the run from the first point to the second? &&&&What therefore is the slope of your line segment? &&&& Between x = 1 and x = 3, do you think the graph of the actual function f(x) is a straight line? Why or why not? &&&& Your line segment went from one point of the f(x) graph to another. A line segment from one point on a curve to another is called a secant. So you have just calculated the slope of a secant line. Restate this in your own words: &&&& Your secant is a straight line segment, and you have just found its slope. What are the coordinates of the x = 1 point? &&&& You have found the slope of the secant, and you have just given the coordinates of a point on that secant. How do you find the equation of a straight line when you know its slope and the coordinates of a point on the line? &&&& What therefore is the equation of the secant line? &&&& Now repeat the same reasoning to do the following: • Find the x = 1 and x = 2 points of the graph. The x = 1 point will be the same as before. Then find the slope of the secant line between the x = 1 and x = 2 points. Finally find the equation of the secant line and simplify it to the form y = m x + b. Having gone through the process for these points, you should obtain the equation y = -5 x + 4. Now, since the present function is f(x) = x - 2 x^2, the point (x, f(x)) has coordinates ( x, x - 2 x^2). What were the coordinates of your x = 1 point again? &&&& Between your x = 1 point and the general point (x, x - 2 x^2): What are the two x coordinates of your points? (The coordinates of the second point both contain x's). &&&& What therefore is the expression for the run between these points? (there will be x's in this expression also) &&&& What are the two y coordinates of your points? (yes, there will be x's in this expression .. ) &&&& What therefore is the expression for the slope between these points? (and it will be no surprise to find x's in this expression) &&&& A line segment from your x = 1 point to the point (x, x - 2 x^2) is a secant of the graph of f(x) = x - 2 x^2. You have just found the slope of the secant. Explain this in your own words: &&&& Your expression for the secant included the variable x. If you replace x by 2 and evaluate your expression for the secant, what do you get? &&&& What were the coordinates of your x = 1 point again? &&&& What was the expression you got for the slope of the secant line when you replaced x by 2? &&&& What therefore is the equation of the secant line through your x = 1 point, when x has been replaced by 2? &&&& What does all this tell you about the given solution to the problem? &&&& If you had trouble understanding the second half of the given solution, you should consider submitting the optional exercise. If you do, insert each answer in the line that contains the &&&&, but before the &&&& mark, so that your insertion ends with &&&&. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.3.36 / 50 (was 3.2.40). h(x) = 3 x^3 + 5. Is the function even, odd or neither? How did you determine algebraically that this is the case? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: h(x) = 3x^3 +5 h(-x) = 3-x^3 +5 h = -3x^3 + 5 function is not even and the function is not odd confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * h(x) = 3x^3 +5 h(-x) = 3-x^3 +5 = -3x^3 + 5 h(x) is not equal to h(-x), which means that the function is not even. h(x) is not equal to -h(-x), since 3 x^3 + 5 is not equal to - ( -3x^3 + 5) = 3 x^3 - 5. So the function is not odd either. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth 158 024. `* 24*********************************************
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Given Solution: * * A linear function, represented most simply by y = x, has no curvature. A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down. A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity. A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change. A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote. An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape. The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'. The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.4.20 (was 3.3.12). Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My sketch increases, and it has a decreasing rate. However, the points on my sketch aren’t exactly right on the dot, so I’m not sure which points they are exactly. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4. sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2. The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x. So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. ** What three points did you label on your graph? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2x + 5 y intercept: 5 slope:2 x intercept: 2x + 5 = 0 or x = -5/2 y = 2x + 5 straight line: (-3, -1), (-5/2, 0) and (0, 5) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2x + 5 y intercept: 5 slope:2 x intercept: 2x + 5 = 0 or x = -5/2 y = 2x + 5 straight line: (-3, -1), (-5/2, 0) and (0, 5) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2x + 5 y intercept: 5 slope:2 x intercept: 2x + 5 = 0 or x = -5/2 y = 2x + 5 straight line: (-3, -1), (-5/2, 0) and (0, 5) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&! ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 2x + 5 y intercept: 5 slope:2 x intercept: 2x + 5 = 0 or x = -5/2 y = 2x + 5 straight line: (-3, -1), (-5/2, 0) and (0, 5) confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!#*&!